Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, k\neq 0$. $\dfrac{{(r^{-3}k^{4})^{-3}}}{{(r^{-3}k^{5})^{3}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(r^{-3}k^{4})^{-3} = (r^{-3})^{-3}(k^{4})^{-3}}$ On the left, we have ${r^{-3}}$ to the exponent ${-3}$ . Now ${-3 \times -3 = 9}$ , so ${(r^{-3})^{-3} = r^{9}}$ Apply the ideas above to simplify the equation. $\dfrac{{(r^{-3}k^{4})^{-3}}}{{(r^{-3}k^{5})^{3}}} = \dfrac{{r^{9}k^{-12}}}{{r^{-9}k^{15}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{9}k^{-12}}}{{r^{-9}k^{15}}} = \dfrac{{r^{9}}}{{r^{-9}}} \cdot \dfrac{{k^{-12}}}{{k^{15}}} = r^{{9} - {(-9)}} \cdot k^{{-12} - {15}} = r^{18}k^{-27}$